---
title: "Force and position on inclined door"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_3/fluid_mechanics:door-force-and-position"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/door-force-and-position
site: BCT Engineering
---

# Force and position on inclined door

Fluid Mechanics · Chapter 3: Hydrostatics

- Asked in: 2078 Kartik, 10 marks
- Video: [Fluid Mechanics Exam Problem: Hydrostatic Forces on a Plane Gate](https://www.youtube.com/watch?v=31lijQM469k) — Fluid Matters
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/door-force-and-position
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3

## Question

Determine the force and its position from fluids acting on the door as shown in figure. [Figure: closed tank with air gauge pressure 800 kPa above water 1.5 m deep, connected to an inclined door inclined at 60° with the horizontal; door length 3 m; door full-size view is a circle of radius R = 3 m]

## Short answer

The hydrostatic force on the circular door is5849 kN (5.85×10³ kN). Its centre of pressure lies5.8 mm below the centroid along the slope, nearly coincident due to the large air pressure.

## Answer

The figure is only partly legible. **Assumed reading:** a circular door of diameter 3 m (radius $R=1.5$ m, length 3 m along the slope), inclined at $\theta=60^\circ$ to the horizontal, with its upper edge 1.5 m below the water surface. The air above the water has gauge pressure $p_0=800$ kPa. $g=9.81$.

```text
   air 800 kPa
  ---------------- water surface
   1.5 m
    \
     \  door 3 m along slope
      \   60 deg
```

**Depth of centroid:**

$$
\bar h=1.5+R\sin60^\circ=1.5+1.5(0.866)=2.799\ \text{m}
$$

**Pressure at the centroid:**

$$
p_c=p_0+\gamma\bar h=800+9.81(2.799)=827.46\ \text{kPa}
$$

**Area:** $A=\pi R^{2}=\pi(1.5)^{2}=7.069\ \text{m}^2$.

**Force:**

$$
F=p_cA=827.46\times7.069=5849\ \text{kN}
$$

**Position.** The air pressure is equivalent to an extra head $p_0/\gamma=81.55$ m of water above the surface. Distance of the centroid from this imaginary free surface along the slope:

$$
\bar y=\frac{81.55+2.799}{\sin60^\circ}=97.40\ \text{m}
$$

$$
\bar y_{cp}-\bar y=\frac{I_G}{A\bar y}=\frac{\pi R^{4}/4}{\pi R^{2}\,\bar y}=\frac{R^{2}}{4\bar y}=\frac{2.25}{4(97.40)}=0.0058\ \text{m}
$$

**Answer: $F\approx5.85\times10^{3}$ kN (5849 kN), acting normal to the door at about 5.8 mm below the centre of the door, measured along the slope.** The large air pressure makes the centre of pressure almost coincide with the centroid.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the total force on the plane door found?
   A. F = pressure at the centroid x area, with p_c = p0 + γ h
   B. F = p0 x area only
   C. F = γ h x area only
   D. F = pressure at the top edge x area
   Answer: A. Force on a plane surface is the centroid pressure times area; here the air pressure is included. (Key point: Force on the door)

2. How is the 800 kPa air pressure treated when locating the centre of pressure?
   A. It is ignored for the position
   B. It acts at the door's top edge
   C. As an extra head p0/γ = 81.55 m of water above the surface (imaginary free surface)
   D. It is subtracted from the water depth
   Answer: C. Measure ȳ from the equivalent free surface, which gives a very large ȳ. (Key point: Handling the air pressure)

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